MCQ
$\frac{\sec 49^{\circ}}{\operatorname{cosec} 41^{\circ}}=$ ?
- ✓$1$
- B$0$
- C$\frac{3}{4}$
- DNone of these
✓
Answer
Correct option: A.
$1$
$\frac{\sec 49^{\circ}}{\operatorname{cosec} 41^{\circ}}=\frac{\sec \left(90^{\circ}-41^{\circ}\right)}{\operatorname{cosec} 41^{\circ}}$
$=\frac{\operatorname{cosec} 41^{\circ}}{\operatorname{cosec} 41^{\circ}} (\because \sec (90-\theta)=\operatorname{cosec} \theta)$
$=1$
$=\frac{\operatorname{cosec} 41^{\circ}}{\operatorname{cosec} 41^{\circ}} (\because \sec (90-\theta)=\operatorname{cosec} \theta)$
$=1$
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